Answer
$x= 2 $, $x= 38 $
Work Step by Step
Given \begin{equation}
\sqrt{2 x+5}-3=\sqrt{x-2}.
\end{equation} Start by squaring both sides to eliminate the radical sign on the right hand side. Rearrange the equation and then square both sides to eliminate the radical sign on the left hand side and then solve for $x$.
\begin{equation}
\begin{aligned}
\sqrt{2 x+5}-3&=\sqrt{x-2}\\
\left(\sqrt{2 x+5}-3\right)^2&= \left(\sqrt{x-2}\right)^2\\
2 x+5-6\sqrt{2 x+5}+9& = x-2 \\
-6\sqrt{2 x+5}& =-(x+16)\\
\left( 6\sqrt{2 x+5}\right)^2 & = \left( x+16\right)^2\\
72x+180& =x^2+32x+256\\
0& =x^2-40x+76.
\end{aligned}
\end{equation} Use the quadratic formula to find the value(s) of $x$. \begin{equation}
\begin{aligned}
a &= 1\ , b= -40\ ,\ c = 76\\
x & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
x&=\frac{-(-40) \pm \sqrt{(-40)^2-4 \cdot 1 \cdot(76)}}{2 \cdot 1} \\
&=\frac{40\pm 36}{2}\\
&= 20\pm 18\\
\implies x& = 38\\
x& = 2.
\end{aligned}
\end{equation} Check.
\begin{equation}
\begin{aligned}
\sqrt{2 \cdot 2+5}-3&\stackrel{?}{=}\sqrt{2-2} \\
0& =0\quad \checkmark\\
\sqrt{2 \cdot 38+5}-3&\stackrel{?}{=}\sqrt{38-2} \\
6& =6\quad \checkmark\\
\end{aligned}
\end{equation} The solution is $x= 2 $, $x= 38 $.
